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The potential of hydrogen electrode having a pH=10 isA. 0.59VB. 0.00VC. `-0.59V`D. `-0.059V` |
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Answer» Correct Answer - C `H^(+)+erarr(1)/(2)H_(2)` `E=E^(@)-(0.0591)/(1)"log"(1)/([H^(+)])=-0.0591xxpH` `=0-0.0591xx10=-0.591V`. |
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